The discussion revolves around solving the equation ##y=\mathrm{exp}(\frac{-x\pi}{\sqrt{1-x^2}})## for x when y = 0.1, which involves concepts from exponential functions and logarithms.
There are multiple interpretations being explored, with participants providing suggestions for improving the equation's presentation and considering different algebraic manipulations. Guidance has been offered without a clear consensus on the next steps.
Participants are working under the constraints of a homework assignment, which may limit the information available or the methods that can be used. The original poster's attempts are noted, but no explicit resolution has been reached.
It is all right so far. What is your problem? You got a simple equation for x2.Captain1024 said:Homework Statement
Solve ##y=\mathrm{exp}(\frac{-x\pi}{\sqrt{1-x^2}})## for x when y = 0.1
Homework Equations
##\mathrm{ln}(e^x)=x##
The Attempt at a Solution
##\mathrm{ln}(0.1)=\frac{-x\pi}{\sqrt{1-x^2}}##
##(\frac{-\mathrm{ln}(0.1)}{\pi})^2=\frac{x^2}{1-x^2}##
Captain1024 said:Homework Statement
Solve ##y=\mathrm{exp}(\frac{-x\pi}{\sqrt{1-x^2}})## for x when y = 0.1
Homework Equations
##\mathrm{ln}(e^x)=x##
The Attempt at a Solution
##\mathrm{ln}(0.1)=\frac{-x\pi}{\sqrt{1-x^2}}##
##(\frac{-\mathrm{ln}(0.1)}{\pi})^2=\frac{x^2}{1-x^2}##
##x=\sqrt{\frac{1}{\frac{\pi ^2}{(\mathrm{ln}(10))^2}+1}}##SammyS said:One thing that will improve the looks of the right hand side. Recognize that −ln(0.1) is ln( (0.1)−1 )
Another thing you might try is to take the reciprocal of both sides of the equation.